When a line or a curve is approaching a given line or curve that too very closely, then the line or the curve that is approaching is termed as an asymptote. In other words we can say that the distance between the given curve and the asymptote approaches to zero as they both are tending to infinity.

We can have three kinds of asymptotes for any given function:

1) Vertical asymptotes

2) Horizontal asymptotes

3) Oblique asymptotes

We say that a line x = c to be a vertical asymptote of a function y = f (x) only if either $lim_{x -> c-}$ $f (x)$ = $\pm$ $\infty$ or $lim_{x -> c+}$ f (x) = $\pm$ $\infty$. 

The horizontal lines where the function’s graph is approaching as x approaches to $\pm$ $\infty$ are called horizontal asymptotes.

When we have a linear asymptote which is neither horizontal nor vertical that is not parallel to any axis, then we call it a slant or oblique asymptote.


There is no any fixed formula to find the asymptotes of a function. To understand this concept, consider a rational function y = $\frac{f (x)}{g (x)}$, given g (x) is not equal to zero.
Then the vertical asymptotes of 'y' can be determined by substituting the denominator equal to zero, that is, g (x) = 0. This will give all possible vertical asymptotes.

Now to find horizontal asymptotes we check the degree of the functions f (x) and g (x). If the degree of f (x) is less than the degree of g (x) then x – axis is the horizontal asymptote that is y = 0. When the degrees of f (x) and g (x) are same then the horizontal asymptote is given by dividing the coefficients of leading terms of f (x) and g (x). If the degree of f (x) is greater than the degree of g (x) then we do not have a horizontal asymptote.

Slant asymptotes only exist in functions where the degree of f (x) is greater than the degree of g (x). In such case we divide f (x) by g (x), and the polynomial part or the quotient so obtained (neglecting the remainder), say q (x) is the slant asymptote written as y = q (x).


Example 1: Find all asymptotes of the function below.

y = $\frac{(x^2 + 5 x + 6)}{(9 x^2 -4)}$

Solution: Let us take denominator equal to zero which will give us vertical asymptote.

$\rightarrow$ 9$x^2$ - 4 = 0

$\rightarrow$9 $x^2$ = 4

$\rightarrow$ $x^2$ = $\frac{4}{9}$

$\rightarrow$ x = $\pm$ ($\frac{2}{3}$)

Here the degrees of the numerator and the denominator are same. Hence the horizontal asymptote can be found by dividing the coefficients of the

leading terms, that is y = $\frac{x^2}{9 x^2}$ = $\frac{1}{9}$

So we can recollect all asymptotes here as below.

Vertical asymptotes: x = $\frac{2}{3}$, x = -$\frac{2}{3}$

Horizontal asymptote: y = $\frac{1}{9}$

Slant asymptote: None.

Example 2: Determine the slant asymptote below.

y = $\frac{(x^2 + 3)}{(x + 9)}$

Solution: Here the degree of numerator is greater than that of denominator so slant asymptote exist.

When we divide $x^2$ + 3 by $x$ + 9 we get (x - 9) + $\frac{ 84}{(x + 9)}$

Hence the slant asymptote here is y = x – 9.